求助PREP一题~~~~~~~~~~~~~~~~~~~~~

rx_11 · 发表于 2010-7-16 14:06:06

if x and y are positive integers, is x an even integer?
(1) x(y+5) is an even integer.
(2) 6y^2+41y+25 is an even integer
答案是E。。不解···求解答~~~

youngerlee · 发表于 2010-7-16 14:18:51

with following steps, I get an E answer.....

if x and y are positive integers, is x an even integer?
(1) x(y+5)=even
(2) 6y^2+41y+25=even

[1]x(y+5)=even

when y is even, y+5 is an odd, then x should be even.
then y is odd, y+5 is even, we have no idea about x in this case.

insufficient.

[2]6y^2+41y+25=even

25 odd, so 6y^2+41y should be odd AND SHOULD NOT BE 0

6y^2+41y should be odd means y[6y+41]should be odd

so y and [6y+41] should be all odds

when y even, 6y+41 odd y[6y+41] even
when y odd, 6y+41 odd y[6y+41] odd

then, y must be odd

take 'y=odd' into factor [1]
when y odd, x have no idea

that's the answer

insufficient

rx_11 · 发表于 2010-7-16 14:23:54

我也很想说：哪尼！！！！！

NN快来啊··

youngerlee · 发表于 2010-7-16 14:27:39

我也很想说：哪尼！！！！！

NN快来啊··

-- by 会员 rx_11 (2010/7/16 14:23:54)

soli, 刚才做错一步,现在做出来确实是E了

rx_11 · 发表于 2010-7-16 14:28:42

我也很想说：哪尼！！！！！

NN快来啊··

-- by 会员 rx_11 (2010/7/16 14:23:54)

soli, 刚才做错一步,现在做出来确实是E了

-- by 会员 youngerlee (2010/7/16 14:27:39)

why??

youngerlee · 发表于 2010-7-16 14:35:14

我也很想说：哪尼！！！！！

NN快来啊··

-- by 会员 rx_11 (2010/7/16 14:23:54)

soli, 刚才做错一步,现在做出来确实是E了

-- by 会员 youngerlee (2010/7/16 14:27:39)

why??

-- by 会员 rx_11 (2010/7/16 14:28:42)

见2楼

wenjie87 · 发表于 2010-7-16 14:36:56

（1）X（Y+5）偶
当Y为奇数时，Y+5为偶数，所以无论X是奇还是偶，算式都为偶
（2）6y^2+41y+25 is an even integer
6Y^2为偶
25为奇
6Y^2+25为奇数，只有41Y为奇数时，等式才能是偶数
所以Y为奇数
综上，选E

毛毛carina · 发表于 2010-7-16 14:38:00

条件二是可以推出y是奇数没有错

但把y为继续代入条件一，即y+5为偶数时，x不管是奇是偶，结果都是满足x（y+5）为偶数的，所以两个条件加起来还是insufficient

rx_11 · 发表于 2010-7-16 14:40:08

...
understand~
thank you~

求助PREP一题~~~~~~~~~~~~~~~~~~~~~

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